If a straight line perpendicular to 2x - 3y + | vedclass

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(B) The given line is $2x - 3y + 7 = 0$. Any line perpendicular to the given line is of the form $3x + 2y + k = 0$. To find the intercepts,set $y = 0$

Vedclass · Accepted Answer

$3x + 2y = \pm 6$

Vedclass · Answer

(B) The given line is $2x - 3y + 7 = 0$. Any line perpendicular to the given line is of the form $3x + 2y + k = 0$. To find the intercepts,set $y = 0$: $3x + k = 0 \Rightarrow x = -\frac{k}{3}$. Set $x = 0$: $2y + k = 0 \Rightarrow y = -\frac{k}{2}$. The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x_{intercept} \cdot y_{intercept}| = 3$. $\frac{1}{2} |(-\frac{k}{3}) \cdot (-\frac{k}{2})| = 3$. $\frac{1}{2} |\frac{k^2}{6}| = 3$. $|k^2| = 36 \Rightarrow k = \pm 6$. Substituting $k$ into the equation $3x + 2y + k = 0$,we get $3x + 2y = \pm 6$.

If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

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